Sunday, March 6, 2022

Biggest k such that n over k is less than or equal to n!

Let's find the biggest k such that $n^k \le n! $

In an introductory class for the induction method, one of the common examples given by the teachers is "prove that $n^n \ge n!$". This problem is intuitive since both of the sides are calculated doing n different multiplications while in $n!$ the multipliers are generally less than n. 

While the teacher was proving this in the whiteboard, I came up with a simple question: What is the biggest possible value for k such that $n^k \le n! $. At first I wanted k and n to be an integers but the question can be solved in a similar manner if k is a positive real number. You should try to solve it on your own but here's my solution:

Take the logarithm of both sides with your favorite base:

\[\ln{n^k} \le \ln{n!} \]

\[k \, \ln{n} \le \ln{n} +\ln{n-1}+....+ \ln{2} + \ln{1} \]

\[ k \le \frac{ \ln{n} +\ln{n-1}+....+ \ln{2} + \ln{1} }{\ln{n}} \]

Round k to the biggest possible integer such that this holds.

As you can see when we plot the numbers it's a pretty linear graph but when we look at the ratio between n and k the graph changes:

And when we look at the limit using wolframalpha, just like you can see from the graph, n/k converges to 1.

You can find this sequence in OEIS with the code number A039960.


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